I am currently stuck on how to find the radius as shown in my xy-projection it is cut by the plane x=1. However I tried ignoring that part as I cannot find a solution for it and first tried to find an answer without that given plane. I arrived at this answer.$ \int_0^ \frac{\pi}{3} \int_0^2 \int_0^ \frac{\pi}{3} rdzdrd\theta $
How will I be able to determine the radius if it's projection is shaped like so? Any insight would be appreciated.
The region is in first octant, bound by planes $y = 0, x = 1$ and $y = \sqrt3 \, x$ and is bound below and above by paraboloids $z = x^2 + y^2$ and $z = 8 - x^2 - y^2$ respectively.
So the integral is pretty straightforward if we set up in the order $dz$ first. In cartesian coordinates,
$\displaystyle \int_0^{1} \int_0^{x \sqrt3} \int_{x^2+y^2}^{8 - x^2 - y^2} dz \, dy \, dx$
In cylindrical coordinates,
$\displaystyle \int_0^{\pi/3} \int_0^{1/cos \theta} \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta$
As you may have already observed, paraboloids intersect each other at $z = 4$, the plane $y = \sqrt3 \, x$ cuts the paraboloids between $0 \leq z\ \leq 8$ but the plane $x = 1$ cuts the paraboloids only between $z = 1$ and $z = 7$. So $x, y$ is bound by paraboloid and the plane $y = \sqrt3 \, x$ for $0 \leq z \leq 1, 7 \leq z \leq 8$. Between $1 \leq z \leq 7$, it has both parts where $x, y$ are bound by the plane(s) or is bound by the paraboloids. So trying to integrate over $dz$ last will make the integral complicated and you will have to split it (or use min function). But if you are interested in seeing it, let me know.